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M1 OCR (Not MEI) Exam - 9/06/2015

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Reply 100
Avatar for kawehi
kawehi
3
Original post by MsFahima
Check the opening post please.. is there any other important tips I should add? Thanks! :biggrin:


I would definitely talk about these:

Taut/inextensible string means that tension is constant throughout the string.
ALWAYS remember to work in components (horizontal/vertical) (parallel/perpendicular) NEVER just start randomly adding forces together
Draw out the entire diagram, writing down everything you know
Don't make any assumptions!

I'll let you know if I think of any more :smile: And I didn't realise that you were putting links to all of the answers, that's a really good idea!!
Reply 101
Avatar for MsFahima
MsFahima
OP
20
Original post by kawehi
I would definitely talk about these:

Taut/inextensible string means that tension is constant throughout the string.
ALWAYS remember to work in components (horizontal/vertical) (parallel/perpendicular) NEVER just start randomly adding forces together
Draw out the entire diagram, writing down everything you know
Don't make any assumptions!

I'll let you know if I think of any more :smile: And I didn't realise that you were putting links to all of the answers, that's a really good idea!!


Thanks! I'll add that to the opening post. They are some good points. :smile:
Reply 102
Avatar for kawehi
kawehi
3
Original post by Super199
ah I see. So I shouldn't assume it will be the same as the previous part.

The next question is 4iia.
For the part above that I worked out the speed of L after the collision to be 1ms^-1.

It then says M collides with N. From looking at the mark scheme it changes back to using 2ms^-1. Why is that?
If you could explain that part it would be appreciated.

Thanks :smile:


(a) Find the total momentum of M and N in the direction of M’s motion before this collisiontakes place

Pay attention to the question! :smile: The direction of motion is always considered positive!
Original post by kawehi
(a) Find the total momentum of M and N in the direction of M’s motion before this collisiontakes place

Pay attention to the question! :smile: The direction of motion is always considered positive!


But isn't M going to the right now?
Because the collision of L and M has made it change direction?

If it was considering M as it was before surely they wouldn't collide?
(edited 8 years ago)
,alallallalal.pngCan anyone help me with the last part about the contact forces please as I don't understand the mark scheme
Reply 105
Avatar for kawehi
kawehi
3
Original post by Super199
But isn't M going to the right now?
Because the collision of L and M has made it change direction?

If it was considering M as it was before surely they wouldn't collide?


Yeah, M is going towards the right, but now its asking you to consider the motion AFTER the collision!
You can see this by the statement:

(ii) M then collides with N

Just think of the question in chronological order!
Reply 106
Avatar for MsFahima
MsFahima
OP
20
Original post by Fandalf
,alallallalal.pngCan anyone help me with the last part about the contact forces please as I don't understand the mark scheme


What paper is it?

EDIT: Basically, you have to get the resultant of the forces acting on the particle. So when P is in motion, the resultant of Fr and R would be the contact force acting on it. And when the particle is stationary, the only component is mg. And you have to find the angle at which it is from the plane. That's what I've understood from the mark scheme.
(edited 8 years ago)
Original post by kawehi
Yeah, M is going towards the right, but now its asking you to consider the motion AFTER the collision!
You can see this by the statement:

(ii) M then collides with N

Just think of the question in chronological order!


oh ffs, I read the question wrong. It says M's speed remains at 2ms^-1. That makes sense now
Thanks :smile:

How do I do 5iii?
Reply 108
Avatar for kawehi
kawehi
3
Original post by Super199
oh ffs, I read the question wrong. It says M's speed remains at 2ms^-1. That makes sense now
Thanks :smile:

How do I do 5iii?


I did it using SUVAT, because we can find the u, and know a and t.
Just consider the segment from t=4 to t=7, first.

s= ?
u= 0.9(42) = 14.4
v=
a= 7.2
t= 3

then you can work that out from there, right? Just solve for s, then add it onto the total displacement for t=0 to t=4.
Reply 109
Avatar for kawehi
kawehi
3
Original post by Fandalf
,alallallalal.pngCan anyone help me with the last part about the contact forces please as I don't understand the mark scheme


My initial guess would be that it's just 0.6gsin30. If that's right, I'd be happy to explain it! Could you let me know the answer if it isn't, btw :smile:
Original post by kawehi
I did it using SUVAT, because we can find the u, and know a and t.
Just consider the segment from t=4 to t=7, first.

s= ?
u= 0.9(42) = 14.4
v=
a= 7.2
t= 3

then you can work that out from there, right? Just solve for s, then add it onto the total displacement for t=0 to t=4.


ah I see. I'll probably be back later. Thanks for the help :smile:
Reply 111
Avatar for MsFahima
MsFahima
OP
20
Original post by kawehi
My initial guess would be that it's just 0.6gsin30. If that's right, I'd be happy to explain it! Could you let me know the answer if it isn't, btw :smile:


That's not the answer.

Here's the marks scheme:

http://www.ocr.org.uk/Images/61835-mark-scheme-unit-4728-mechanics-1-january.pdf

:smile:
Reply 112
Avatar for kawehi
kawehi
3
Original post by MsFahima


Hmmm, that's weird..

So, I get that the mark scheme has found the vertical component (weight) and the fact that it's 60 degrees from the slope, but how does that answer the question they asked? Contact forces are always perpendicular to the surface, right?? I'm confused haha
Reply 113
Avatar for kawehi
kawehi
3
Original post by Super199
ah I see. I'll probably be back later. Thanks for the help :smile:


No problem!
Reply 114
Avatar for MsFahima
MsFahima
OP
20
Original post by kawehi
Hmmm, that's weird..

So, I get that the mark scheme has found the vertical component (weight) and the fact that it's 60 degrees from the slope, but how does that answer the question they asked? Contact forces are always perpendicular to the surface, right?? I'm confused haha


Yeah, that's what I though but I guess they're asking for the magnitude. And there isn't any videos explaining it as well ... :frown:
Reply 115
Avatar for kawehi
kawehi
3
Original post by MsFahima
Yeah, that's what I though but I guess they're asking for the magnitude. And there isn't any videos explaining it as well ... :frown:


Ok, I'm going to work through the problem and perhaps I'll come back with an explanation in about 5 mins *fingers crossed*
Reply 116
Avatar for MsFahima
MsFahima
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20
Original post by kawehi
Ok, I'm going to work through the problem and perhaps I'll come back with an explanation in about 5 mins *fingers crossed*


Sure! I did the first part.. it was fine. I'm not sure about that second part.. Tell me if you figure it out!
Original post by MsFahima
Sure! I did the first part.. it was fine. I'm not sure about that second part.. Tell me if you figure it out!



Original post by kawehi
Ok, I'm going to work through the problem and perhaps I'll come back with an explanation in about 5 mins *fingers crossed*


Are you guys doing C3/C4?
Reply 118
Avatar for MsFahima
MsFahima
OP
20
Original post by Super199
Are you guys doing C3/C4?


I am. You?
Original post by MsFahima
I am. You?


Yup. How is revision going for them?

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